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satela [25.4K]
1 year ago
8

The magnetic field 40.0 cm away from a long, straight wire carrying current 2.00 A is 1.00μT. (a) At what distance is it 0.100μ

Physics
1 answer:
maw [93]1 year ago
4 0

The distance should be 4m from the wire in order to get the magnetic field of 0.100μ .

  • The magnitude and direction of the magnetic field due to a straight  wire carrying current can be calculated using the previously mentioned Biot-Savart law. Let "I" be the current flowing in a straight line and "r" be the distance. Then the magnetic field produced by the wire at that particular point is given by  B=\frac{u_0I}{2\pi r}  ...(1)
  • Since the wire is assumed to be very long, the magnitude of the magnetic field depends on the distance of the point from the wire rather than the position along the wire.

It is given that magnetic field 40.0 cm away from a straight wire is  1.00μT having  current 2.00 A .

From equation (1)  magnetic field 40.0 cm = 0.4m away from a straight wire is 1.00μT which is given by    1.00=\frac{u_0I}{2\pi \times0.4}      .....(2)

From equation (1)  magnetic field 'r' m away from a straight wire is 0.100μT which is given by    0.100=\frac{u_0I}{2\pi \times r}       ...(3)

On dividing equation (2) by (3) , we get

             \frac{1}{0.1} =\frac{r}{0.4} \\\\r=4m

Learn more about magnetic field here :

brainly.com/question/27939568

#SPJ4

 

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Answer:

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Explanation:

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(\frac{V_{1}}{V_{2}})^{\gamma -1} = \frac{T_{2}}{T_{1}}.

Now, let's use the ideal gas equation to the initial and the final state:

\frac{p_{1} V_{1}}{T_{1}} = \frac{p_{2} V_{2}}{T_{2}}

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Combining this equation with the first equation we have:

(\frac{T_{1}p_{2}}{T_{2}p_{1}})^{\gamma -1} = \frac{T_{2}}{T_{1}}

(\frac{p_{2}}{p_{1}})^{\gamma -1} = \frac{T_{2}^{\gamma}}{T_{1}^{\gamma}}

Now, we just need to solve this equation for T₂.

T_{1}\cdot (\frac{p_{2}}{p_{1}})^{\frac{\gamma - 1}{\gamma}} = T_{2}

Let's assume the initial temperature and pressure as 25 °C = 298 K and 1 atm = 1.01 * 10⁵ Pa, in a normal conditions.

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Finally, T2 will be:

T_{2}=278.80 K

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